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Answer:
See explanation below
Explanation:
The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.
Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.
For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)
For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.
When using ion-selective electrodes, to compensate for a complex or unknown matrix, the standard addition method can be used to determine the analyte concentration. Option D
<h3>What are ion-selective electrodes?</h3>
Analytical chemistry is a science that deal with the measurement and detection of the accurate amount of a substance. Analytical chemistry plays a large role in environmental management as it helps in the determination of the levels of contaminants in a sample.
An ion selective electrode is used in analytical chemistry to measure the amount of a target ion by converting its activity into a measurable electrical signal.
Hence, when using ion-selective electrodes, to compensate for a complex or unknown matrix, the standard addition method can be used to determine the analyte concentration.
Learn more about ion-selective electrodes:brainly.com/question/14987024
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Answer:
The correct answer is 0.047 mol/L
Explanation:
The atmospheric air is a mixture of gases. We can assume an ideal behavior of the gas and use the ideal gas equation:
PV= nRT
where P is the pressure, V is the volume, n is the number of moles, R is a constant (0.082 L.atm/K.mol) and T is the temperature in K.
We have to first convert the pressure from Torr to atm:
760 Torr= 1 atm
⇒ 718 Torr x 1 atm/760 Torr = 0.945 atm
Then, we convert the temperature from ºC to K:
0ºC = 273 K
⇒ -29ºC+273= 244 K
Finally, we introduce the data in the equation and calculate de densitiy, which is the moles per liters of gas (n/V):
PV = nRT
n/V= P/RT
n/V = (0.945 atm)/(0.082 L.atm/K.mol x 244 K) = 0.047 mol/L